MHB Show the following sequence as a monotone increasing

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The sequence defined by the terms ${x}_{n}$ in set $S$ is confirmed to be monotone increasing due to the condition ${x}_{n-1} < x_n$ for all $n \ge 2$. This strict inequality satisfies the definition of monotonicity, making further proof unnecessary. The supremum of the set $S$ does not affect the monotonicity of the sequence. Participants in the discussion agree that the initial conditions provided are sufficient for the conclusion. Therefore, the sequence is indeed monotone increasing as stated.
cbarker1
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Dear Everyone, Here is the sequence: Let $S\subset\Bbb{R}$ and ${x}_{n}\in S$ and $S\ne\emptyset$ . ${x}_{n-1}<{x}_{n}\le\sup S$ for all $n\ge2$. Prove the sequence is monotone increasing.

I need help proving it; I do not know where to start? Thanks
Carter
 
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I'm confused: Isn't this already given since $x_{n-1} < x_n$ for all $n \ge 2$, assuming $x_1$ is the first term in the sequence?
 
The assumption is correct where ${x}_{1}$ is given. So I can say the sequence is monotone increasing?
 
Yes. I mean, you are given terms $x_n$ in a set $S$ such that $x_{n-1} < x_n$ for all $n \ge 2$.
That latter inequality is the definition of strict monotonicity, so if the exercise really reads like this, then I cannot see what you would have to do else. (The supremum does not play any role, either.)

Anyone else here on board that sees something I overlooked?
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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